CN101293261A - Support roll for flattener - Google Patents

Abstract

The invention discloses a supporting roll used in a straightening machine, including a supporting roll set used for supporting a straightener roll; the roll shape of the middle section of the body of the supporting roll is a quartic or quadratic polynomial curve along the diametric direction; the invention can eliminate the stress concentration occurring in the contact process of the roll set, relieve the local abrasion of the supporting roll, prolong the service life of the roll set and achieve the purposes of prolonging on-machine time, reducing shutdown polishing time and lowering production cost.

Description

A kind of backing roll that is used for straightener

Technical field

The present invention relates to a kind of straightener backing roll, relate in particular to the thermal straightening machine backing roll that Wide and Heavy Plates is used.

Background technology

At present, be used for the thermal straightening machine equipment of Wide and Thick Slab production in the world based on quadruple 9 roller reversible straighteners.The aligning thickness range is 10 ~ 100mm, and the straightening temperature of steel plate is generally between 400 ~ 900 ℃.Steel plate is general aligning 1 ~ 3 passage on thermal straightening machine, and the yield strength under the steel plate of being rectified is hot is 1000Mpa to the maximum.Aligning speed is determined according to the thickness of steel plate, velocity interval 0-2.5m/s.Have roller, laterally fascinate, vertically fascinate, go into/outlet roller regulates separately and dynamic function such as roll gap adjusting, can eliminate the various flatness defects of steel plate, but also can align clapboard, satisfy user's requirement better, smoothing roll adopts the internal water cooling cooling, can whole fast roll changing.Adopt all-hydraulic roll gap Adjustment System, and be furnished with advanced process control technology and straightener model control technology, realized full automatic working, have very high setting accuracy, improved leveling effect.

Present topmost problem is the roll marks problem.Former multi-roll straightening machine technology is shown in Fig. 1 ~ 3, upper straightening roll 1 is hung at upper frame 3 by stacking spring 4, the backing roll 2 of upper straightening roll 1 then is fastened on the upper frame 3 by bearing block 5 and bolt 6, and backing roll 2 roll shapes are generally formed (as shown in Figure 4) by plain-barreled roll edged portion's chamfering or middle plain-barreled roll roll shape edged portion's tapering 7 and axle head circular arc fillet 8.Because being rigidly connected of top backing up roll 2 bearing blocks and upper frame 3 is difficult to guarantee that all top backing up roll 2 lower surfaces are in same plane.And, owing to the piecewise function that consists of of backing roll roll shape, can only guarantee that first order derivative is continuous, and second derivative is discontinuous.Therefore, the contact condition that will guarantee top backing up roll 2 and upper straightening roll 1 in the practical application is unusual difficulty, and the stress of average contact stress more than 3 times can occur being higher than at limit portion tapering 7 and middle plain-barreled roll transition position and concentrate.This machining accuracy that has caused occurring all backing roll diameters must guarantee very strict, the very harsh requirement in the 0.05mm scope, however also is difficult to avoid occurring the roll marks problem.

Characteristics according to backing roll and smoothing roll; if will solve because the roll marks problem that causes of loose contact, can from by improve backing roll or smoothing roll roll surface material, improve rotary resistance that the backing roll hermetically-sealed construction reduces backing roll, a kind of online reconditioning instrument of design with reduce downtime, by structure and the roll shape methods such as eliminating loose contact fully that matches.

Patent 200510011792.3 provides a kind of repair weld composite leveller roll, provide resurfacing welding material and bead-welding technology, had advantages such as body of roll intensity height, antifatigue, shock resistance be strong, corrosion-resistant, wear-resistant and low cost of manufacture, life-span is long, can realize repeatedly reparative regeneration.This patent mainly is at smoothing roll, has improved every performance of smoothing roll, but can not solve above-mentioned roll marks problem.

Patent 200420090448.9 has provided a kind of online reconditioning instrument of smoothing roll, at the smoothing roll impression that occurs in producing with cohere.But, when the problematic roll surface of needs reconditioning, must shut down, influenced operating efficiency thus.

Summary of the invention

The purpose of this invention is to provide a kind of backing roll that is used for straightener,, avoided generation, thereby reach the purpose that to improve roller be service life, reduce downtime in backing roll body of roll end and the local contact phenomena of smoothing roll by optimizing the backing roll roller curve.

The object of the present invention is achieved like this:

A kind of backing roll that is used for straightener comprises the backing roll roller group that is used to support a smoothing roll, and the roll shape of this backing roll roll body center section part is a fourth-degree polynomial curve.

The equation of described fourth-degree polynomial curve is:

$y=\underset{i=0}{\overset{4}{\mathrm{\Σ}}}{a}_i{x}^i$

In the formula,

$a_0=m_1{a}_{\mathrm{<figure-callout\; id="0"\; label="q"\; filenames="A20071003983700282.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}0}^{2/3}+m_{2}B(L^3-12{\mathrm{<figure-callout\; id="2"\; label="LB"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">LB</figure-callout>}}^2+{6B}^3)$

$a_1=m_1[m_3{a}_{q0}{a}_{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}1}+\frac{{\mathrm{<figure-callout\; id="4"\; label="m"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">m</figure-callout>}}_4}{27}(27a_{q3}{a}_{q6}-{9a}_{q4}{a}_{q5}-10\frac{a_{\mathrm{<figure-callout\; id="5"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}5}^3}{a_{q6}})]+m_2\frac{B}{L}(12B^3-8B^{2}L+L^3)$

$a_2=m_1[m_3(a_{q0}{a}_{q2}-\frac{1}{6}{a}_{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}1}^2)+m_4(a_{q4}{a}_{q6}-\frac{1}{6}{a}_{\mathrm{<figure-callout\; id="5"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}5}^2)]+6m_{2}B(2B-L)$

$a_3=m_1[m_4{a}_{q5}{a}_{\mathrm{<figure-callout\; id="6"\; label="q"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}6}+\frac{{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}_3}{27}(27a_{q0}{a}_{q3}-{9a}_{q1}{a}_{q2}-10\frac{a_{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}1}^3}{a_{q0}})]+{2m}_2(2B-L)$

$a_4=m_1{a}_{\mathrm{<figure-callout\; id="6"\; label="q"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}6}^{2/3}+{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}_2$

Wherein,

${\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}_1={\left(\frac{9{\mathrm{\&Delta;x}}^2}{16{\mathrm{RE}}^{*2}}\right)}^{\frac{1}{3}}$

${\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}_2=\frac{q^{*}}{24E_2{I}_2}$

${\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}_3=\frac{1}{3}{\left(a_{q0}\right)}^{-\frac{4}{3}}$

${\mathrm{<figure-callout\; id="4"\; label="m"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">m</figure-callout>}}_4=\frac{1}{3}{\left(a_{q6}\right)}^{-\frac{4}{3}}$

Wherein,

a _q0＝q ₀+q ₁+q ₃

a _q1＝q ₁m+q ₂-q ₃m+q ₄

$a_{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}2}=\frac{1}{2}{q}_1{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^2+q_{2}m+\frac{1}{2}{q}_3{m}^2-q_{4}m$

$a_{\mathrm{<figure-callout\; id="3"\; label="q"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}3}=\frac{1}{6}{q}_1{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3+\frac{1}{2}{q}_2{m}^2-\frac{1}{6}{q}_3{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3+\frac{1}{2}{q}_4{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$

$a_{\mathrm{<figure-callout\; id="4"\; label="q"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">q</figure-callout>}4}=\frac{1}{24}{q}_1{\mathrm{<figure-callout\; id="4"\; label="m"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">m</figure-callout>}}^4+\frac{1}{6}{q}_2{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3+\frac{1}{24}{q}_3{m}^4-\frac{1}{6}{q}_4{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3$

$a_{\mathrm{<figure-callout\; id="5"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}5}=\frac{1}{120}{q}_1{\mathrm{<figure-callout\; id="5"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^5+\frac{1}{24}{q}_2{m}^4-\frac{1}{120}{q}_3{\mathrm{<figure-callout\; id="5"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^5+\frac{1}{24}{q}_4{\mathrm{<figure-callout\; id="4"\; label="m"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">m</figure-callout>}}^4$

$a_{\mathrm{<figure-callout\; id="6"\; label="q"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}6}=\frac{1}{120}{q}_2{m}^5-\frac{1}{120}{q}_4{\mathrm{<figure-callout\; id="5"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^5$

Wherein,

${\mathrm{<figure-callout\; id="0"\; label="q"\; filenames="A20071003983700282.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}}_0=\frac{E_2{I}_2}{E_1{\mathrm{<figure-callout\; id="1"\; label="I"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">I</figure-callout>}}_1+E_2{I}_2}(\frac{F}{L-2B}+\frac{1}{2}p)$

q ₁＝g ₈+g ₉q ₂

${\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}}_2=\frac{g_3-{\mathrm{mg}}_8{e}^{g_{10}}+{\mathrm{<figure-callout\; id="6"\; label="mg"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">mg</figure-callout>}}_6+g_4(\frac{1}{2}{g}_{10}+1)}{({\mathrm{<figure-callout\; id="9"\; label="mg"\; filenames="A20071003983700292.png"\; state="\{\{state\}\}">mg</figure-callout>}}_9+\frac{1}{2}{g}_{10}-1)e^{g_{10}}-{\mathrm{mg}}_7-g_5(\frac{1}{2}{g}_{10}+1)}$

q ₃＝g ₆+g ₇q ₂

q ₄＝g ₄+g ₅q ₂

Wherein,

${\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_1=\frac{2m}{D_{\mathrm{\&tau;}}}(\frac{\mathrm{<figure-callout\; id="2"\; label="F"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">F</figure-callout>}}{2\mathrm{\&pi;}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_1^2(L-2B)G_1}-D_{\mathrm{\&tau;}}{\mathrm{<figure-callout\; id="0"\; label="q"\; filenames="A20071003983700282.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}}_0+\frac{M_0}{E_1{I}_1})$

${\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2=-\frac{p(L-2B)-2M_1}{D_{\mathrm{\&tau;}}{E}_2{\mathrm{<figure-callout\; id="2"\; label="I"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">I</figure-callout>}}_2}$

${\mathrm{<figure-callout\; id="3"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_3=\frac{{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^2}{2}(L-2B)(p-q_0)+{\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2$

${\mathrm{<figure-callout\; id="4"\; label="g"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">g</figure-callout>}}_4=-\frac{1}{2}{g}_3{e}^{\frac{1}{2}{g}_{10}}$

${\mathrm{<figure-callout\; id="5"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_5=-e^{g_{10}}$

${\mathrm{<figure-callout\; id="6"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_6=\frac{1}{2m}(g_1-{\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2+\frac{{\mathrm{<figure-callout\; id="3"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_3}{2}{e}^{\frac{1}{2}{g}_{10}})$

${\mathrm{<figure-callout\; id="7"\; label="g"\; filenames="A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_7=\frac{1-3{\mathrm{<figure-callout\; id="5"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_5}{2m}$

$g_8=\frac{1}{2m}({\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2+\frac{{\mathrm{<figure-callout\; id="3"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_3}{2}{e}^{\frac{1}{2}{g}_{10}})$

${\mathrm{<figure-callout\; id="9"\; label="g"\; filenames="A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_9=\frac{3-{\mathrm{<figure-callout\; id="5"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_5}{2m}$

g ₁₀＝m(L-2B)

Wherein,

$m=\sqrt{\frac{2(\frac{1}{E_1{\mathrm{<figure-callout\; id="1"\; label="I"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">I</figure-callout>}}_1}+\frac{1}{E_2{I}_2})}{D_{\mathrm{\&tau;}}}}$

p＝αF _L/W _b

$I_1=\frac{\mathrm{\&pi;}}{4}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_1^4$

$I_2=\frac{\mathrm{\&pi;}}{4}{\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_2^4$

$D_{\mathrm{\&tau;}}=\frac{1}{{\mathrm{\&pi;}\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_1^2{\mathrm{<figure-callout\; id="1"\; label="G"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">G</figure-callout>}}_1}+\frac{1}{{\mathrm{\&pi;<figure-callout\; id="2"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_2^2{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">G</figure-callout>}}_2}$

M ₁＝FB

$R={(\frac{1}{R_1}+\frac{1}{R_2})}^{-1}$

$E^{*}={(\frac{1-v_1^2}{{\mathrm{<figure-callout\; id="1"\; label="E"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">E</figure-callout>}}_1}+\frac{1-v_2^2}{E_2})}^{-1}$

$q^{*}=\frac{1}{L-2B}{\&Integral;}_0^{L-2B}q\left(x\right)\mathrm{dx}$

Wherein,

M ₀The additional bending moment of having ignored the smoothing roll other parts and produce except that the roll body center section part can be found the solution according to the statically indeterminate beam analytical method;

F is the support reaction of backing roll maximum, can utilize the mechanical balance equation solution to obtain by test or according to design feature;

G ₁, E ₁, G ₂, E ₂, be respectively modulus of shearing, the elastic modelling quantity of smoothing roll and backing roll material, unit is N/m2;

R ₁, R ₂It is the radius of smoothing roll and backing roll;

L is a top backing up roll two bearings centre distance;

B is the distance of back-up roll bearing center to edge, homonymy contact zone;

F _LIt is the straightener load;

W _bIt is steel plate width;

α is the unbalance loading coefficient, can determine according to position (represent concrete backing roll 2 and smoothing roll 1 contact area) and the magnitude of load that bear of backing roll on the smoothing roll length direction, is generally 1.0 ~ 13.6;

The size of the micro unit that the backing roll length direction was divided when Δ x was calculating;

X is the coordinate along the horizontal axis barrel length.

Preferably, the end face of described backing roll roll body adopts fillet to be connected with roll shape along diametric curve.

The roll shape of the center section part of described backing roll roll body is the quadratic polynomial curve.

Described quadratic polynomial curvilinear equation is:

$y=\underset{i=0}{\overset{2}{\mathrm{\&Sigma;}}}{a}_i{x}^i$

In the formula,

$a_0=m_1{a}_{\mathrm{<figure-callout\; id="0"\; label="q"\; filenames="A20071003983700282.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}0}^{2/3}+m_{2}B(L^3-12{\mathrm{<figure-callout\; id="2"\; label="LB"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">LB</figure-callout>}}^2+{6B}^3)$

$a_1=m_1[m_3{a}_{q0}{a}_{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}1}+\frac{{\mathrm{<figure-callout\; id="4"\; label="m"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">m</figure-callout>}}_4}{27}(27a_{q3}{a}_{q6}-{9a}_{q4}{a}_{q5}-10\frac{a_{\mathrm{<figure-callout\; id="5"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}5}^3}{a_{q6}})]+m_2\frac{B}{L}(12B^3-8B^{2}L+L^3)$

$a_2=m_1[m_3(a_{q0}{a}_{q2}-\frac{1}{6}{a}_{\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}1}^2)+m_4(a_{q4}{a}_{q6}-\frac{1}{6}{a}_{\mathrm{<figure-callout\; id="5"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}5}^2)]+6m_{2}B(2B-L)$

Wherein,

${\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}_1={\left(\frac{9{\mathrm{\&Delta;x}}^2}{16{\mathrm{RE}}^{*2}}\right)}^{\frac{1}{3}}$

${\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}_2=\frac{q^{*}}{24E_2{\mathrm{<figure-callout\; id="2"\; label="I"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">I</figure-callout>}}_2}$

${\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}_3=\frac{1}{3}{\left(a_{q0}\right)}^{-\frac{4}{3}}$

${\mathrm{<figure-callout\; id="4"\; label="m"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">m</figure-callout>}}_4=\frac{1}{3}{\left(a_{q6}\right)}^{-\frac{4}{3}}$

Wherein,

a _q0＝q ₀+q ₁+q ₃

a _q1＝q ₁m+q ₂-q ₃m+q ₄

$a_{\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}2}=\frac{1}{2}{q}_1{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^2+q_{2}m+\frac{1}{2}{q}_3{m}^2-q_{4}m$

$a_{\mathrm{<figure-callout\; id="3"\; label="q"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}3}=\frac{1}{6}{q}_1{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3+\frac{1}{2}{q}_2{m}^2-\frac{1}{6}{q}_3{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3+\frac{1}{2}{q}_4{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$

$a_{\mathrm{<figure-callout\; id="4"\; label="q"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">q</figure-callout>}4}=\frac{1}{24}{q}_1{\mathrm{<figure-callout\; id="4"\; label="m"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">m</figure-callout>}}^4+\frac{1}{6}{q}_2{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3+\frac{1}{24}{q}_3{m}^4-\frac{1}{6}{q}_4{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">m</figure-callout>}}^3$

$a_{\mathrm{<figure-callout\; id="6"\; label="q"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}6}=\frac{1}{120}{q}_2{m}^5-\frac{1}{120}{q}_4{\mathrm{<figure-callout\; id="5"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^5$

Wherein,

${\mathrm{<figure-callout\; id="0"\; label="q"\; filenames="A20071003983700282.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}}_0=\frac{E_2{I}_2}{E_1{I}_1+E_2{I}_2}(\frac{F}{L-2B}+\frac{1}{2}p)$

q ₁＝g ₈+g ₉q ₂

${\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">q</figure-callout>}}_2=\frac{g_3-{\mathrm{mg}}_8{e}^{g_{10}}+{\mathrm{<figure-callout\; id="6"\; label="mg"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">mg</figure-callout>}}_6+g_4(\frac{1}{2}{g}_{10}+1)}{({\mathrm{<figure-callout\; id="9"\; label="mg"\; filenames="A20071003983700292.png"\; state="\{\{state\}\}">mg</figure-callout>}}_9+\frac{1}{2}{g}_{10}-1)e^{g_{10}}-{\mathrm{mg}}_7-g_5(\frac{1}{2}{g}_{10}+1)}$

q ₃＝g ₆+g ₇q ₂

q ₄＝g ₄+g ₅q ₂

Wherein,

${\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_1=\frac{2m}{D_{\mathrm{\&tau;}}}(\frac{\mathrm{<figure-callout\; id="2"\; label="F"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">F</figure-callout>}}{2\mathrm{\&pi;}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_1^2(L-2B)G_1}-D_{\mathrm{\&tau;}}{\mathrm{<figure-callout\; id="0"\; label="q"\; filenames="A20071003983700282.png,A20071003983700292.png"\; state="\{\{state\}\}">q</figure-callout>}}_0+\frac{M_0}{E_1{I}_1})$

${\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2=-\frac{p(L-2B)-2M_1}{D_{\mathrm{\&tau;}}{E}_2{\mathrm{<figure-callout\; id="2"\; label="I"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">I</figure-callout>}}_2}$

${\mathrm{<figure-callout\; id="3"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_3=\frac{{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">m</figure-callout>}}^2}{2}(L-2B)(p-q_0)+{\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2$

${\mathrm{<figure-callout\; id="4"\; label="g"\; filenames="A20071003983700251.png"\; state="\{\{state\}\}">g</figure-callout>}}_4=-\frac{1}{2}{g}_3{\mathrm{<figure-callout\; id="1"\; label="e"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">e</figure-callout>}}^{\frac{1}{2}{g}_{10}}$

${\mathrm{<figure-callout\; id="5"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_5=-e^{g_{10}}$

${\mathrm{<figure-callout\; id="6"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_6=\frac{1}{2m}(g_1-{\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2+\frac{{\mathrm{<figure-callout\; id="3"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_3}{2}{\mathrm{<figure-callout\; id="1"\; label="e"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">e</figure-callout>}}^{\frac{1}{2}{g}_{10}})$

${\mathrm{<figure-callout\; id="7"\; label="g"\; filenames="A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_7=\frac{1-3{\mathrm{<figure-callout\; id="5"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_5}{2m}$

$g_8=\frac{1}{2m}({\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_2+\frac{{\mathrm{<figure-callout\; id="3"\; label="g"\; filenames="A20071003983700251.png,A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_3}{2}{\mathrm{<figure-callout\; id="1"\; label="e"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">e</figure-callout>}}^{\frac{1}{2}{g}_{10}})$

${\mathrm{<figure-callout\; id="9"\; label="g"\; filenames="A20071003983700292.png"\; state="\{\{state\}\}">g</figure-callout>}}_9=\frac{3-{\mathrm{<figure-callout\; id="5"\; label="g"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">g</figure-callout>}}_5}{2m}$

g ₁₀＝m(L-2B)

Wherein,

$m=\sqrt{\frac{2(\frac{1}{E_1{I}_1}+\frac{1}{E_2{I}_2})}{D_{\mathrm{\&tau;}}}}$

p＝αF _L/W _b

$I_1=\frac{\mathrm{\&pi;}}{4}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_1^4$

$I_2=\frac{\mathrm{\&pi;}}{4}{\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_2^4$

$D_{\mathrm{\&tau;}}=\frac{1}{{\mathrm{\&pi;}\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_1^2{\mathrm{<figure-callout\; id="1"\; label="G"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">G</figure-callout>}}_1}+\frac{1}{{\mathrm{\&pi;<figure-callout\; id="2"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_2^2{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">G</figure-callout>}}_2}$

M ₁＝FB

$R={(\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">R</figure-callout>}}_1}+\frac{1}{R_2})}^{-1}$

$E^{*}={(\frac{1-{\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">v</figure-callout>}}_1^2}{{\mathrm{<figure-callout\; id="1"\; label="E"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">E</figure-callout>}}_1}+\frac{1-{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="A20071003983700251.png,A20071003983700322.png"\; state="\{\{state\}\}">v</figure-callout>}}_2^2}{E_2})}^{-1}$

$q^{*}=\frac{1}{L-2B}{\&Integral;}_0^{L-2B}q\left(x\right)\mathrm{dx}$

Wherein,

M ₀Owing to having ignored the additional bending moment that smoothing roll 1 other parts produce, can find the solution according to the statically indeterminate beam analytical method;

F is the support reaction of backing roll maximum, can utilize the mechanical balance equation solution to obtain by test or according to design feature;

G ₁, E ₁, G ₂, E ₂, be respectively modulus of shearing, the elastic modelling quantity of smoothing roll and backing roll material, unit is N/m2;

R ₁, R ₂It is the radius of smoothing roll and backing roll;

L is a top backing up roll two bearings centre distance;

B is the distance of back-up roll bearing center to edge, homonymy contact zone;

F _LIt is the straightener load;

W _bIt is steel plate width;

α is the unbalance loading coefficient, can determine according to position (represent concrete backing roll 2 and smoothing roll 1 contact area) and the magnitude of load that bear of backing roll on the smoothing roll length direction, is generally 1.0 ~ 13.6;

The size of the micro unit that the backing roll length direction was divided when Δ x was calculating;

X is the coordinate along the horizontal axis barrel length.

Preferably, the end face of described backing roll roll body adopts fillet to be connected with roll shape along diametric curve.

The present invention has improved the backing roll roller curve; having eliminated roller is that the stress that occurs in the contact process is concentrated; alleviate the concentrated wear of backing roll, prolonged the service life of roller system, reached the purpose that improves its floor-to-floor time, reduces the shutdown reconditioning time, reduces production costs.

Description of drawings

Fig. 1 is a slab thermal straightening machine schematic diagram;

Fig. 2 is whole assembly relation schematic diagram for the slab thermal straightening machine shows top roll;

Fig. 3 is the front view of Fig. 2;

Fig. 4 shows the schematic diagram of top backing up roll roll shape for former slab thermal straightening machine;

Fig. 5 is the schematic diagram of the utility model top backing up roll roll shape;

Fig. 6 is the distribution of contact trend contrast of the new and old roll shape of 0mm when the roller value for first embodiment;

Fig. 7 is the distribution of contact trend contrast of the new and old roll shape of 1mm when the roller value for first embodiment;

Fig. 8 is the distribution of contact trend contrast of the new and old roll shape of 2mm when the roller value for first embodiment;

Fig. 9 is the distribution of contact trend contrast of the new and old roll shape of 3mm when the roller value for first embodiment;

Figure 10 is the maximum contact stress contrast of the new and old roll shape of first embodiment;

Figure 11 is the high degree contrast of the urgency of the new and old roll shape of first embodiment;

Figure 12 is that second embodiment is the distribution of contact trend contrast of the new and old roll shape of 0mm when the roller value;

Figure 13 is that second embodiment is the distribution of contact trend contrast of the new and old roll shape of 1mm when the roller value;

Figure 14 is that second embodiment is the distribution of contact trend contrast of the new and old roll shape of 2mm when the roller value;

Figure 15 is that second embodiment is the distribution of contact trend contrast of the new and old roll shape of 3mm when the roller value;

Figure 16 is the maximum contact stress contrast of second new and old roll shape of embodiment;

Figure 17 is the high degree contrast of the urgency of second new and old roll shape of embodiment;

Figure 18 is that the 3rd embodiment is the distribution of contact trend contrast of the new and old roll shape of 0mm when the roller value;

Figure 19 is that the 3rd embodiment is the distribution of contact trend contrast of the new and old roll shape of 1mm when the roller value;

Figure 20 is that the 3rd embodiment is the distribution of contact trend contrast of the new and old roll shape of 2mm when the roller value;

Figure 21 is that the 3rd embodiment is the distribution of contact trend contrast of the new and old roll shape of 3mm when the roller value;

Figure 22 is the maximum contact stress contrast of the 3rd the new and old roll shape of embodiment;

Figure 23 is the high degree contrast of the urgency of the 3rd the new and old roll shape of embodiment.

The specific embodiment

Embodiment 1:

Shown in Fig. 2,3,5, one cover slab thermal straightening machine, every upper straightening roll 1 has 6 backing rolls 2, these 6 backing rolls 2 are numbered into 1 ', 2 ', 3 ', 4 ', 5 ', 6 ', wherein 1 ' and 6 ' number backing roll is similar operating condition, can compile is one group, and in like manner 2 ' and 5 ' is one group, and 3 ' and 4 ' is one group.Maximum smoothing roll diameter is φ 360mm, and the smoothing roll diameter of bore is φ 180mm (a logical cooling water), and the maximal support roller diameter is φ 370mm, and backing roll length is 500mm, and there is the round-corner transition of 5mm in backing roll body of roll limit portion.Total straightening force is 2000t to the maximum.Roller curve is relative plain-barreled roll, is that basic grinding forms with plain-barreled roll, so is roller footpath maximum in the middle of the backing roll, is φ 370mm.

According to theory analysis, and cooperate corresponding result of finite element, determine that the backing roll roll shape is a fourth-degree polynomial curve.According to the backing roll present position, consider the full payload size and the unbalance loading form of bearing, determine corresponding unbalance loading coefficient, promptly

3 ' and 4 ' number backing roll, α=4.2

2 ' and 5 ' number backing roll, α=2.9

1 ' and 6 ' number backing roll, α=3.7

Adopt identical roller curve form and different roll shape parameters, the reference axis round dot schedules the backing roll geometric center:

3 ' and 4 ' number backing roll, the semidiameter distribution curve is

y＝2.728×10 ^-2+1.5371×10 ^-4(251-x)+1.1815×10 ^-5(251-x) ²

-9.2554×10 ^-10(251-x) ³+5.3433×10 ^-10(251-x) ⁴

y＝2.728×10 ^-2+1.5371×10 ^-4(251+x)+1.1815×10 ^-5(251+x) ²

-9.2554×10 ^-10(251+x) ³+5.3433×10 ^-10(251+x) ⁴

2 ' and 5 ' number backing roll, the semidiameter distribution curve is

y＝1.9464×10 ^-2+1.1558×10 ^-4(251-x)+1.2743×10 ^-5(251-x) ²

-6.6254×10 ^-10(251-x) ³+3.8604×10 ^-10(251-x) ⁴

y＝1.9464×10 ^-2+1.1558×10 ^-4(251+x)+1.2743×10 ^-5(251+x) ²

-6.6254×10 ^-10(251+x) ³+3.8604×10 ^-10(251+x) ⁴

1 ' and 6 ' number backing roll, the semidiameter distribution curve is

y＝2.4316×10 ^-2+1.3949×10 ^-4(251-x)+1.2442×10 ^-5(251-x) ²

-8.2556×10 ^-10(251-x) ³+4.7885×10 ^-10(251-x) ⁴

y＝2.4316×10 ^-2+1.3949×10 ^-4(251+x)+1.2442×10 ^-5(251+x) ²

-8.2556×10 ^-10(251+x) ³+4.7885×10 ^-10(251+x) ⁴

Wherein, x ∈ [0,246].For the fillet design of backing roll, should choose within the specific limits according to different actual requirements.The fillet design principle is as follows, and the minimum of a value of fillet should be greater than according to roller being the fillet value that the contact strength requirement is determined.Consider that roller system wearing and tearing back roller curve changes, the roller end contact stress will increase.Therefore, the roller when being plain-barreled roll according to backing roll is the contact strength and the backing roll strength of materials limit, and principle is that fillet should guarantee that the touch-safe coefficient of backing roll is greater than 5.The maximum of fillet should be less than according to roller being the fillet value that wear range is determined between the contact zone and during machine.The normal roll change cycle of thermal straightening machine is long, mostly is 1 year twice most, and the wearing and tearing of roller system cannot be avoided.Therefore, in design during fillet, should be taken into account that in roll change cycle roller is the variation of the contact range brought of the situation of change of wear extent and roller curve, principle is radius of corner＜barrel length-(contact length of roller system after the greatest wear+20mm)/2; For guaranteeing that enough floor-to-floor times are arranged, consider the quickening of rhythm of production simultaneously, fillet is chosen unsuitable excessive, answers as close as possible lower limit; Fillet should guarantee simultaneously tangent with roller curve and roller end face.

The effect of present embodiment is intermediate support roller (3 ' and 4 ' number backing roll) state among the figure shown in Fig. 5 ~ 11.Init is an original roller type, and new is new roll shape.The implication of anxious high degree is a certain distribution is compared in this numerical value reflection with normal distribution sharp-pointed degree or a smooth degree.Positive peak is represented sharp-pointed relatively distribution.Negative peak is represented the distribution of relatively flat.

The new roll shape of backing roll is with the increase of roller, stress remains on an acceptable level, maximum stress and contact width change not obvious, and concentrated of stress occurs when maximum roller, specific stress is bigger than normal mutually but with old roll shape, contact width reduces to some extent, and contact area moves to the straightener center.New roll shape is compared with old roll shape, influence significantly reduces roller to maximum stress, being roller does not have old roll shape obvious to the influence of contact condition, and anxious high degree all is a negative value, this illustrates that new roll shape contact condition is more stable, the adaptive capacity that external condition is changed is stronger, and has eliminated the stress concentrated position.

Embodiment 2:

Shown in Fig. 2,3,5, every upper straightening roll 1 has 6 backing rolls 2, according to the straightener design feature then 1 ' and 6 ' number backing roll be similar operating condition, can compile is one group, in like manner 2 ' and 5 ' is one group, 3 ' and 4 ' is one group.Smoothing roll is a solid roll, and maximum smoothing roll diameter is φ 360mm, and the maximal support roller diameter is φ 370mm, and backing roll length is 500mm, and there is the round-corner transition of 5mm in backing roll body of roll limit portion.Total straightening force is 2000t to the maximum.Roller curve is relative plain-barreled roll, is that basic grinding forms with plain-barreled roll, so is roller footpath maximum in the middle of the backing roll, is φ 370mm.

According to former theoretical analysis, and cooperate corresponding result of finite element, through calculating a that finds in the roller curve equation ₃x ³And a ₄x ⁴Value very little, be the 1.0E-20 order of magnitude to the maximum, thus the decision omit.Therefore determine that the backing roll roll shape is the quadratic polynomial form.According to the backing roll present position, adopt identical roller curve form and different roll shape parameters, the reference axis round dot schedules backing roll left end face center:

3 ' and 4 ' number backing roll, the semidiameter distribution curve is

y＝3.2×10 ^-3x ²-4.6807×10 ^-7x+1.096×10 ^-5，x∈[5，250]

y＝3.2×10 ^-6(500-x) ²-4.6807×10 ^-10(500-x)+1.096×10 ^-8，x∈[250，495]

2 ' and 5 ' number backing roll, the semidiameter distribution curve is:

y＝1.6×10 ^-3x ²-5.8525×10 ^-8x+1.3791×10 ^-6，x∈[5，250]

y＝1.6×10 ^-3(500-x) ²-5.8525×10 ^-8(500-x)+1.3791×10 ^-6，x∈[250，495]

1 ' and 6 ' number backing roll, the semidiameter distribution curve is

y＝1.6×10 ^-3x ²-5.8525×10 ^-8x+1.3791×10 ^-6，x∈[5，250]

y＝1.6×10 ^-3(500-x) ²-5.8525×10 ^-8(500-x)+1.3791×10 ^-6，x∈[250，495]

Effect of the present utility model is intermediate support roller (3 ' and 4 ' number backing roll) state among the figure shown in Figure 12 ~ 17.

Embodiment 3:

3 backing roll supportings are arranged on the every smoothing roll, and backing roll length is 1386mm, and other condition is with embodiment 1.

By the distribution form and the carrying size of backing roll, extract according to branch

The intermediate support roller, α=10.6

About two backing rolls, α=5.8

According to the backing roll present position, adopt identical roller curve form and different roll shape parameters, the reference axis round dot schedules the backing roll geometric center:

The intermediate support roller, the semidiameter distribution curve is

y＝-1.3113×10 ^-3+8.2481×10 ^-6(695.772-x)+5.7355×10 ^-06(695.772-x) ²

-2.0182×10 ^-12(695.772-x) ³+2.2430×10 ^-11(695.772-x) ⁴

y＝-1.3113×10 ^-3+8.2481×10 ^-6(695.772+x)+5.7355×10 ^-06(695.772+x) ²

-2.0182×10 ^-12(695.772+x) ³+2.2430×10 ^-11(695.772+x) ⁴

About two backing rolls, the semidiameter distribution curve is

y＝-9.1951×10 ^-4+5.4588×10 ^-6(695.772-x)+3.8220×10 ^-06(695.772-x) ²

-1.1290×10 ^-12(695.772-x) ³+1.4736×10 ^-11(695.772-x) ⁴

y＝-9.1951×10 ^-4+5.4588×10 ^-6(695.772+x)+3.8220×10 ^-06(695.772+x) ²

-1.1290×10 ^-12(695.772+x) ³+1.4736×10 ^-11(695.772+x) ⁴

Wherein, x ∈ [0,690.772]

Effect of the present utility model is intermediate support roller state among the figure shown in Figure 18 ~ 23.

Claims (6)

1. a backing roll that is used for straightener comprises the backing roll roller group that is used to support a smoothing roll, it is characterized in that:

The roll shape of backing roll roll body center section part is a fourth-degree polynomial curve.

2. the backing roll that is used for straightener as claimed in claim 1 is characterized in that: described fourth-degree polynomial curve equation is:

$y=\underset{i=0}{\overset{4}{\mathrm{\&Sigma;}}}{a}_i{x}^i$

In the formula,

$a_0=m_1{a}_{q0}^{2/3}+m_{2}B(L^3-12L{B}^2+6B^3)$

$a_1=m_1[m_3{a}_{q0}{a}_{q1}+\frac{m_4}{27}(27a_{q3}{a}_{q6}-9a_{q4}{a}_{q5}-10\frac{a_{q5}^3}{a_{q6}})]+m_2\frac{B}{L}(12B^3-8B^{2}L+L^3)$

$a_2=m_1[m_3(a_{q0}{a}_{q2}-\frac{1}{6}{a}_{q1}^2)+m_4(a_{q4}{a}_{q6}-\frac{1}{6}{a}_{q5}^2)]+6m_{2}B(2B-L)$

$a_3=m_1[m_4{a}_{q5}{a}_{q6}+\frac{m_3}{27}(27a_{q0}{a}_{q3}-9a_{q1}{a}_{q2}-10\frac{a_{q1}^3}{a_{q0}})]+2m_2(2B-L)$

$a_4=m_1{a}_{q6}^{2/3}+m_2$

Wherein,

$m_1={\left(\frac{9\mathrm{\&Delta;}{x}^2}{16R{E}^{*2}}\right)}^{\frac{1}{3}}$

$m_2=\frac{q^{*}}{24E_2{I}_2}$

$m_3=\frac{1}{3}{\left(a_{q0}\right)}^{-\frac{4}{3}}$

$m_4=\frac{1}{3}{\left(a_{q6}\right)}^{-\frac{4}{3}}$

Wherein,

a _q0＝q ₀+q ₁+q ₃

a _q1＝q ₁m+q ₂-q ₃m+q ₄

$a_{q2}=\frac{1}{2}{q}_1{m}^2+q_{2}m+\frac{1}{2}{q}_3{m}^2-q_{4}m$

$a_{q3}=\frac{1}{6}{q}_1{m}^3+\frac{1}{2}{q}_2{m}^2-\frac{1}{6}{q}_3{m}^3+\frac{1}{2}{q}_4{m}^2$

$a_{q4}=\frac{1}{24}{q}_1{m}^4+\frac{1}{6}{q}_2{m}^3+\frac{1}{24}{q}_3{m}^4-\frac{1}{6}{q}_4{m}^3$

$a_{q5}=\frac{1}{120}{q}_1{m}^5+\frac{1}{24}{q}_2{m}^4-\frac{1}{120}{q}_3{m}^5+\frac{1}{24}{q}_4{m}^4$

$a_{q6}=\frac{1}{120}{q}_2{m}^5-\frac{1}{120}{q}_4{m}^5$

Wherein,

$q_0=\frac{E_2{I}_2}{E_1{I}_1+E_2{I}_2}(\frac{F}{L-2B}+\frac{1}{2}p)$

q ₁＝g ₈+g ₉q ₂

$q_2=\frac{g_3-m{g}_8{e}^{g_{10}}+m{g}_6+g_4(\frac{1}{2}{g}_{10}+1)}{({\mathrm{mg}}_9+\frac{1}{2}{g}_{10}-1)e^{g_{10}}-{\mathrm{mg}}_7-g_5(\frac{1}{2}{g}_{10}+1)}$

q ₃＝g ₆+g ₇q ₂

q ₄＝g ₄+g ₅q ₂

Wherein,

$g_1=\frac{2m}{D_{\mathrm{\&tau;}}}(\frac{F}{2\mathrm{\&pi;}{R}_1^2(L-2B)G_1}-D_{\mathrm{\&tau;}}{q}_0+\frac{M_0}{E_1{I}_1})$

$g_2=-\frac{p(L-2B)-2M_1}{D_{\mathrm{\&tau;}}{E}_2{I}_2}$

$g_3=\frac{m^2}{2}(L-2B)(p-q_0)+g_2$

$g_4=-\frac{1}{2}{g}_3{e}^{\frac{1}{2}{g}_{10}}$

$g_5=-e^{g_{10}}$

$g_6=\frac{1}{2m}(g_1-g_2+\frac{g_3}{2}{e}^{\frac{1}{2}{g}_{10}})$

$g_7=\frac{1-3g_5}{2m}$

$g_8=\frac{1}{2m}(g_1+g_2+\frac{g_3}{2}{e}^{\frac{1}{2}{g}_{10}})$

$g_9=\frac{3-g_5}{2m}$

g ₁₀＝m(L-2B)

Wherein,

$m=\sqrt{\frac{2(\frac{1}{E_1{I}_1}+\frac{1}{E_2{I}_2})}{D_{\mathrm{\&tau;}}}}$

p＝αF _L/W _b

$I_1=\frac{\mathrm{\&pi;}}{4}{R}_1^4$

$I_2=\frac{\mathrm{\&pi;}}{4}{R}_2^4$

$D_{\mathrm{\&tau;}}=\frac{1}{\mathrm{\&pi;}{R}_1^2{G}_1}+\frac{1}{\mathrm{\&pi;}{R}_2^2{G}_2}$

M ₁＝FB

$R={(\frac{1}{R_1}+\frac{1}{R_2})}^{-1}$

$E^{*}={(\frac{1-v_1^2}{E_1}+\frac{1-v_2^2}{E_2})}^{-1}$

$q^{*}=\frac{1}{L-2B}{\&Integral;}_0^{L-2B}q\left(x\right)\mathrm{dx}$

Wherein,

M ₀Owing to having ignored the additional bending moment that smoothing roll 1 other parts produce, can find the solution according to the statically indeterminate beam analytical method;

F is the support reaction of backing roll maximum, can utilize the mechanical balance equation solution to obtain by test or according to design feature;

G ₁, E ₁, G ₂, E ₂, be respectively modulus of shearing, the elastic modelling quantity of smoothing roll and backing roll material, unit is N/m2;

R ₁, R ₂It is the radius of smoothing roll and backing roll;

L is a top backing up roll two bearings centre distance;

B is the distance of back-up roll bearing center to edge, homonymy contact zone;

F _LIt is the straightener load;

W _bIt is steel plate width;

α is the unbalance loading coefficient, can according to backing roll on the smoothing roll length direction the position and the magnitude of load that bears determine, be generally 1.0 ~ 13.6;

The size of the micro unit that the backing roll length direction was divided when Δ x was calculating;

X is the coordinate along the horizontal axis barrel length.

3. the backing roll that is used for straightener as claimed in claim 2 is characterized in that: the end face of described backing roll adopts fillet to be connected with roll shape along diametric curve.

4. the backing roll that is used for straightener as claimed in claim 1 is characterized in that: the roll shape of supporting roll body center section part is the quadratic polynomial curve.

5. the backing roll that is used for straightener as claimed in claim 4 is characterized in that: described quadratic polynomial curvilinear equation is:

$y=\underset{i=0}{\overset{2}{\mathrm{\&Sigma;}}}{a}_i{x}^i$

In the formula,

$a_0=m_1{a}_{q0}^{2/3}+m_{2}B(L^3-12L{B}^2+6B^3)$

$a_1=m_1[m_3{a}_{q0}{a}_{q1}+\frac{m_4}{27}(27a_{q3}{a}_{q6}-9a_{q4}{a}_{q5}-10\frac{a_{q5}^3}{a_{q6}})]+m_2\frac{B}{L}(12B^3-8B^{2}L+L^3)$

$a_2=m_1[m_3(a_{q0}{a}_{q2}-\frac{1}{6}{a}_{q1}^2)+m_4(a_{q4}{a}_{q6}-\frac{1}{6}{a}_{q5}^2)]+6m_{2}B(2B-L)$

Wherein,

$m_1={\left(\frac{9\mathrm{\&Delta;}{x}^2}{16R{E}^{*2}}\right)}^{\frac{1}{3}}$

$m_2=\frac{q^{*}}{24E_2{I}_2}$

$m_3=\frac{1}{3}{\left(a_{q0}\right)}^{-\frac{4}{3}}$

$m_4=\frac{1}{3}{\left(a_{q6}\right)}^{-\frac{4}{3}}$

Wherein,

α _q0＝q ₀+q ₁+q ₃

a _q1＝q ₁m+q ₂-q ₃m+q ₄

$a_{q2}=\frac{1}{2}{q}_1{m}^2+q_{2}m+\frac{1}{2}{q}_3{m}^2-q_{4}m$

$a_{q3}=\frac{1}{6}{q}_1{m}^3+\frac{1}{2}{q}_2{m}^2-\frac{1}{6}{q}_3{m}^3+\frac{1}{2}{q}_4{m}^2$

$a_{q4}=\frac{1}{24}{q}_1{m}^4+\frac{1}{6}{q}_2{m}^3+\frac{1}{24}{q}_3{m}^4-\frac{1}{6}{q}_4{m}^3$

$a_{q6}=\frac{1}{120}{q}_2{m}^5-\frac{1}{120}{q}_4{m}^5$

Wherein,

$q_0=\frac{E_2{I}_2}{E_1{I}_1+E_2{I}_2}(\frac{F}{L-2B}+\frac{1}{2}p)$

q ₁＝g ₈+g ₉q ₂

$q_2=\frac{g_3-m{g}_8{e}^{g_{10}}+m{g}_6+g_4(\frac{1}{2}{g}_{10}+1)}{({\mathrm{mg}}_9+\frac{1}{2}{g}_{10}-1)e^{g_{10}}-{\mathrm{mg}}_7-g_5(\frac{1}{2}{g}_{10}+1)}$

q ₃＝g ₆+g ₇q ₂

q ₄＝g ₄+g ₅q ₂

Wherein,

$g_1=\frac{2m}{D_{\mathrm{\&tau;}}}(\frac{F}{2\mathrm{\&pi;}{R}_1^2(L-2B)G_1}-D_{\mathrm{\&tau;}}{q}_0+\frac{M_0}{E_1{I}_1})$

$g_2=-\frac{p(L-2B)-2M_1}{D_{\mathrm{\&tau;}}{E}_2{I}_2}$

$g_3=\frac{m^2}{2}(L-2B)(p-q_0)+g_2$

$g_4=-\frac{1}{2}{g}_3{e}^{\frac{1}{2}{g}_{10}}$

$g_5=-e^{g_{10}}$

$g_6=\frac{1}{2m}(g_1-g_2+\frac{g_3}{2}{e}^{\frac{1}{2}{g}_{10}})$

$g_7=\frac{1-3g_5}{2m}$

$g_8=\frac{1}{2m}(g_1+g_2+\frac{g_3}{2}{e}^{\frac{1}{2}{g}_{10}})$

$g_9=\frac{3-g_5}{2m}$

g ₁₀＝m(L-2B)

Wherein,

$m=\sqrt{\frac{2(\frac{1}{E_1{I}_1}+\frac{1}{E_2{I}_2})}{D_{\mathrm{\&tau;}}}}$

p＝αF _L/W _b

$I_1=\frac{\mathrm{\&pi;}}{4}{R}_1^4$

$I_2=\frac{\mathrm{\&pi;}}{4}{R}_2^4$

$D_{\mathrm{\&tau;}}=\frac{1}{\mathrm{\&pi;}{R}_1^2{G}_1}+\frac{1}{\mathrm{\&pi;}{R}_2^2{G}_2}$

M ₁＝FB

$R={(\frac{1}{R_1}+\frac{1}{R_2})}^{-1}$

$E^{*}={(\frac{1-v_1^2}{E_1}+\frac{1-v_2^2}{E_2})}^{-1}$

$q^{*}=\frac{1}{L-2B}{\&Integral;}_0^{L-2B}q\left(x\right)\mathrm{dx}$

Wherein,

M ₀Owing to having ignored the additional bending moment that smoothing roll 1 other parts produce, can find the solution according to the statically indeterminate beam analytical method;

F is the support reaction of backing roll maximum, can utilize the mechanical balance equation solution to obtain by test or according to design feature;

G ₁, E ₁, G ₂, E ₂, be respectively modulus of shearing, the elastic modelling quantity of smoothing roll and backing roll material, unit is N/m2;

R ₁, R ₂It is the radius of smoothing roll and backing roll;

L is a top backing up roll two bearings centre distance;

B is the distance of back-up roll bearing center to edge, homonymy contact zone;

F _LIt is the straightener load;

W _bIt is steel plate width;

α is the unbalance loading coefficient, can be according to position (the expression tool of backing roll on the smoothing roll length direction

The backing roll 2 of body and smoothing roll 1 contact area) and the magnitude of load that bears determine, be generally 1.0 ~ 13.6;

The size of the micro unit that the backing roll length direction was divided when Δ x was calculating;

X is the coordinate along the horizontal axis barrel length.

6. the backing roll that is used for straightener as claimed in claim 5 is characterized in that: the end face of described backing roll adopts fillet to be connected with roll shape along diametric curve.

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